Question

The equilibrium constant $K_p$ for the reaction $2H_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2H_{2}O\left(g\right)$ at 2000 K is $1.6\times {10}^7$. The equilibrium constant of the reaction $H_{2}O\left(g\right)\rightleftharpoons H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$ is ________.

• A. $6.25\times {10}^{-8}$
• B. $2.5\times {10}^{-4}$
• C. $7.5\times {10}^{-12}$
• D. $4\times {10}^{-6}$

Answer 1

The correct option is B $2.5\times {10}^{-4}$

For the reaction, $2H_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2H_{2}O\left(g\right)$,
$K=\frac{{\left[H_{2}O\right]}^2}{{\left[H_2\right]}^2\left[O_2\right]}$
$2H_{2}O\left(g\right)\rightleftharpoons 2H_2\left(g\right)+O_2\left(g\right)$
$K=\frac{{\left[H_2\right]}^2\left[O_2\right]}{[H_{2}O{]}^2}$
$H_{2}O\left(g\right)\rightleftharpoons H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
$K_1=\sqrt{\frac{{\left[H_{2}O\right]}^2}{{\left[H_2\right]}^2\left[O_2\right]}}$
Thus, $K_1=\frac{1}{\sqrt{1.6107}}$$=2.5\times {10}^{-4}$